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Find the wavelength of the light in the fluid. To do this, solve the equation from Part A for .
Express your answer in nanometers, to four significant figures.
ANSWER:
Hint 2. Relating wavelengths and the index of refraction
Recall that the wavelength of light in a medium is related to its wavelength in air by the relation ,
where is the index of refraction for the medium.
ANSWER:
ANSWER:
Understanding Fraunhofer Diffraction
Description: The equations for Fraunhofer diffraction are derived. Then, a simple numerical problem is worked.
Learning Goal:
= 1.329
= 476.2
water ( )
methanol ( )
ethanol ( )
acetone ( )
isopropyl alcohol ( )
saline ( )
To understand the derivations of, and be able to use, the equations for Fraunhofer diffraction.
Diffraction is a general term for interference effects related to edges or apertures. Diffraction is more familiar in waves with longer wavlengths than
those of light. For example, diffraction is what causes sound to bend around corners or spread as it passes through a doorway. Water waves
spread as they pass between rocks near a rugged coast because of diffraction. Two different regimes for diffraction are usually identified: Fresnel
and Fraunhofer.
Fresnel diffraction is the regime in which the diffracted waves are observed close (as compared to the size of the object causing the diffraction) to
the place where they are diffracted. Fresnel diffraction is usually very complicated to work with. The other regime, Fraunhofer diffraction, is much
easier to deal with. Fraunhofer diffraction applies to situations in which the diffracted waves are observed far from the point of diffraction. This
allows a number of simplifying approximations to be used, reducing diffraction to a very manageable problem.
An important case of Fraunhofer diffraction is the pattern formed by light shining through a thin slit onto a distant screen (see the figure).
Notice that if the light from the top of the slit and the light from the bottom of the slit
arrive at a point on the distant screen with a phase difference of , then the electric
field vectors of the light from each part of the slit will cancel completely, resulting in a
dark fringe. To understand this phenomenon, picture a phasor diagram for this scenerio
(as show in the figure).
A phasor diagram consists of vectors (phasors) with magnitude proportional to the
magnitude of the electric field of light from a certain point in the slit. The angle of each
vector is equal to the phase of the light from that point. These vectors are added
together, and the resultant vector gives the net electric field due to light from all points in
the slit. In the situation described above, since the magnitude of the electric field vectors
is the same for light from any part of the slit and the angle of the phasors changes
continuously from to , the phasors will make a complete circle, starting and ending
at the origin. The distance from the origin to the endpoint of the phasor path (also the
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